Homework Help: Potential Gradient [Question(s) regarding the concept]

Tags:
1. Sep 14, 2015

Antonius

[Note from mentor: This was originally posted in a non-homework forum so it doesn't have the homework template.]

-----------------------------------------

Problem:
The surface of a solid metal sphere (radius r = 4.58 cm) is at potential V = 9,851 Volts. Find the magnitude of the potential gradient at distance 9.21r from the center of the sphere (outside the sphere), in V/m.
Assume: V(∞) = 0.

My point: I have this general understanding that the Potential Gradient equals to $\frac { \Delta V } { \Delta x }$ where - $x$ - is the distance between the charge $Q$ and the point (or another charge.)

My question: Well in this problem, I am given $V$ being 9851 Volts but that's on one side.
I watched an educational (physics) video saying that if there is only one charge and no other charges, then the Electric Potential Energy equals to $zero.$ (which makes the Potential (at distance 9.21 * r = 0 V) That's being said, should I just divide the given $V$ by the $distance - x -$ which is also given. [ $9.21 * r$ ]?

Please help me with the concept. If you have a link to an online file that could explain this topic in depth please share it. I really want to learn but what I learned from my textbook is not leading me to find answers to problems I am encountering...

Last edited by a moderator: Sep 14, 2015
2. Sep 14, 2015

DEvens

3. Sep 14, 2015

Antonius

So I did this: $$V = \frac {kq} {r}; 9851 Volts = \frac { 8.9 * 10^9 * q } { 0.0458 m } ; Hence, q = 5.069 C$$
Now since I know the $q$ I decided to find $V$ at $9.1 r$.
To do so I followed these steps: $$V = \frac { 8.9 * 10^9 * 5.069 } { 9.1 * 0.0458 } = 1.08 * 10^11 V$$
My question: Why is that the given Potential is smaller than the potential at $9.1 * r$ ?
Shouldn't it be smaller since $V = \frac { U } { q }$ and $U = \frac {k q} { r }$ ? :/

Last edited: Sep 14, 2015
4. Sep 14, 2015

Staff: Mentor

Double-check your arithmetic in this step.

5. Sep 14, 2015

Antonius

Apparently, I forgot to put a parenthesis while dividing by 8.9e9. jtbell, thank you!

So, q = 5.07e-7
Hence, $\frac {\Delta V} {\Delta r} = -2.36e4 V/m$

I am hoping this answer is correct.

6. Sep 14, 2015

Staff: Mentor

Your q is still off by a factor of 10. Check the decimal points in your input numbers. I used the numbers in the first equation of post #3.

7. Sep 14, 2015

Antonius

Yes, I see. I wrote -7, instead of -8. The final answer that I have gotten is calculated with 5.069*10^(-8). Thank you!

8. Sep 14, 2015

Staff: Mentor

Now for the second part (the gradient), how did you calculate that number, exactly? It's off by (approximately, not exactly) a factor of 10 also.

9. Sep 14, 2015

Antonius

Hmm. What's going on lol.

Okay, let me work on it again.
so if q = 5.069 e -8

Now I need to find V at distance 9.1*r (away from center of sphere)
So V = kq/9.1*r = k*5.069 e -8 / (9.1*0.0458) = 1082.44 Volts

Potential Gradient = delta V / delta x = (1082.44 V - 9851 V) / (9.1*r - r) | 9.1*r - r = r ( 9.1 - 1) = 8.1*r

PG = - 23636.196 V/m = -2.36 e 4 V/m

Am I still wrong?

10. Sep 14, 2015

Staff: Mentor

I would call that the "average gradient" between the two r's. It's analogous to average velocity = (x2 - x1) / (t2 - t1).

The problem asks for the gradient at the second value of r (9.1r). This is analogous to the instantaneous velocity v = dx/dt. As DEvens noted in post #2, "gradient" here means "derivative with respect to r".

11. Sep 14, 2015

Antonius

If $V (at 9.1r) = \frac {kq} {9.1*r}$ then $\frac {dV} {dr} = \frac {kq} {9.1} *(\frac {-1} {r^2})$ here $q = \frac {V_r * r} {k}$ which I plugged in to the equation and got $\frac {dV} {dr} = \frac { -V_r} {9.1*r}$ which makes $\frac {dV} {dr} = -23635.97 = -2.36e10^4$

What's wrong..

Last edited: Sep 14, 2015
12. Sep 14, 2015

Staff: Mentor

Take the derivative before substituting 9.1r (0.4168m) for r. This gives you an equation for the gradient at any r (provided that it's ≥ the radius of the sphere). Then substitute your desired value of r.

13. Sep 14, 2015

Antonius

I did it before subbing a number for r though? And r is not 0.4168 m it's 0.0458 m ( in the question it says its r = 4.58 cm )

14. Sep 14, 2015

Staff: Mentor

When you take the derivative, you need to use an equation with a "generic" r, that can have any value. Your problem statement is confusing because it uses r and 9.1r to refer to "specific" values of r. I would have written it more carefully by calling the radius of the sphere r1 and the other radius r2 = 9.1r1.

15. Sep 14, 2015

Antonius

I got -2.15e5.

Is this what you also got?

Last edited: Sep 14, 2015
16. Sep 14, 2015

Staff: Mentor

That's it.

17. Sep 17, 2015

Alan I

How did you guys get this result because I did it over and over with the values provided and I got something else. Here's a summary of the values used according to my calculations:

ro=0.0458 m
r=9.1r0= 0.41678 m
Q=5.02*10-8 C → from Vsurface
dV/dr=-kq/r2 V/m

⇒dV/dr= (8.99*109) N*m2/C2 * (5.02*10-8) C / (0.41678)2 m2

⇒dV/dr=-2597.36≅-2.60*103 V/m

18. Sep 22, 2015

Antonius

There is a vague step which I did not fully understand at first due to weak math background.

Use "the definition of derivative" and it will be much clear and you will most likely get a correct result